Answer:
315mL
Explanation:
Data obtained from the question include the following:
Molarity of stock solution (M1) = 0.135 M
Volume of stock solution needed (V1) =?
Molarity of diluted solution (M2) = 0.0851 M
Volume of diluted solution (V2) = 500mL
The volume of the stock solution needed can be obtain as follow:
M1V1 = M2V2
0.135 x V1 = 0.0851 x 500
Divide both side by 0.135
V1 = (0.0851 x 500) / 0.135
V1 = 315mL
Therefore, the volume of the stock solution needed is 315mL
Answer:
Sigma bonds: 10
Pi bonds: 4
Explanation:
The compound described must be CH2=CH-CO-CH≡CH. If we look at the compound closely, we will notice that there are 10 sigma bonds and 4 pi bonds.
There are three pi bonds between carbon atoms and one pi bond between a carbon and an oxygen atom (C=O). All these can easily be seen in the structure of the formula chosen in this answer.
Answer:
Using the coarse adjustment knob of the microscope in high power may lead to the breaking of the slide if adjusted and raised the slide too much which can damage the sample as well as the high power lens.
In this case, I would recommend using the fine adjustment knob and moving away from the end of the viewing area of the microscope so there would no collision take place. The fine adjustment will help to get a clear image.
Answer:
A. Similar elements have such different masses that this would lead to a completely random arrangement of elements.
Explanation:
Because it is the number of protons, which is identical to the atomic number, that separates different elements from each other, not the atomic mass.
Because teachers don’t really understand it all that well themselves so they can’t teach it effectively
